The explanation is that:
1. As Vin increases, it makes Vo greater.
2. As Vout gets greater (and since it is fed back to Vn (the inverting input), then a greater voltage is produced at this terminal.
3. As seen before, a greater voltage at Vn will tend to produce a smaller Vo. So this is why Vo goes down. Cause is greater Vn, effect is smaller Vo. This is just how the opamp works internally, which results in reducing Vo.

This is only for opamps configured with negative feedback? Iow, Vo will only go down when Vn is large, if there is negative feedback? Because I recall when starting opamps in the previous page that a large + voltage to Vn (inverting input) would generate a smaller Vo. So its not really a matter of the negative feedback, is it?

Iow, Vo will only go down when Vn is large, if there is negative feedback?

Click to expand...

You're confusing negative feedback with inverting operation.

In the circuit you referenced, the output is connected to the inverting input. If you apply a voltage to the non-inverting terminal and use the Zero Differential Theorem, the voltage applied at the non-inverting terminal must also be at the inverting terminal. Since the output is connected to the inverting terminal, the output must also track the voltage at the non-inverting input.

Yes I understand that if V at Vin (+) is 5V then by Diff Theorem the V at the Vn(-) terminal will also be 5V and then since Vo is connected to Vn(-) will also be 5V. So the first part ( Vin(+) at 5V and Vn(-) at 5V ) happens because of the ideal opamp model without using neg feedback. Then the Vout = Vin (because Vin = Vn) happens because of the negative feedback, right?

So whether the higher voltage at Vn (-) comes from a battery or from the Vout negative-feedback, a greater voltage at the Vn(-) will make Vout go down

The explanation is that:
1. As Vin increases, it makes Vo greater.
2. As Vout gets greater (and since it is fed back to Vn (the inverting input), then a greater voltage is produced at this terminal.
3. As seen before, a greater voltage at Vn will tend to produce a smaller Vo. So this is why Vo goes down. Cause is greater Vn, effect is smaller Vo. This is just how the opamp works internally, which results in reducing Vo.

This is only for opamps configured with negative feedback? Iow, Vo will only go down when Vn is large, if there is negative feedback? Because I recall when starting opamps in the previous page that a large + voltage to Vn (inverting input) would generate a smaller Vo. So its not really a matter of the negative feedback, is it?

Click to expand...

Without negative feedback, it is not possible for Vout to change Vn accordingly.

Yes I understand that if V at Vin (+) is 5V then by Diff Theorem the V at the Vn(-) terminal will also be 5V and then since Vo is connected to Vn(-) will also be 5V. So the first part ( Vin(+) at 5V and Vn(-) at 5V ) happens because of the ideal opamp model without using neg feedback. Then the Vout = Vin (because Vin = Vn) happens because of the negative feedback, right?

So whether the higher voltage at Vn (-) comes from a battery or from the Vout negative-feedback, a greater voltage at the Vn(-) will make Vout go down

Yes I understand that if V at Vin (+) is 5V then by Diff Theorem the V at the Vn(-) terminal will also be 5V and then since Vo is connected to Vn(-) will also be 5V. So the first part ( Vin(+) at 5V and Vn(-) at 5V ) happens because of the ideal opamp model without using neg feedback. Then the Vout = Vin (because Vin = Vn) happens because of the negative feedback, right?

So whether the higher voltage at Vn (-) comes from a battery or from the Vout negative-feedback, a greater voltage at the Vn(-) will make Vout go down

Click to expand...

You are on the right track, but I think you are not quite getting it. Let's look at the behavior of the opamp for finite open loop gain (everything else left as being ideal).

The output of the amp is then governed by

Vo = Av(Vp - Vn)

So what happens if we apply the input signals directly, say by way of batteries or perhaps a couple of potentiometers? Well, then we have

Vo = Av(Vp - Vn)

What if Av = 100?

Then if Vp = 2.000 V and Vn = 2.100 V, Vo would be -10.0 V. While if Vp = 2.100 V and Vn =2.000 V, Vo would be +10.0 V. So just 100 mV of difference would take the magnitude of the output voltage to 10 V, which is likely close to saturation, depending on the power supply rails.

Now let's apply our 2.000 V signal to Vp and connect Vo to Vn. Since Vn = Vo, we now have

Vo = Av(Vp - V0)
Vo(1 + Av) = Av·Vp

Vo = Vp [Av / (1 + Av)]

If Av = 100, then this reduces to

Vo = (100/101) Vp

So if we apply 2.000 V to Vp, the Vo (and Vn) will be 1.980 V. That 20 mV difference is what is producing the 2.000 V at the output. (It would actually be more like a 19.80 mV difference producing the 1.980 V at the output.)

Now let's look at what happens if something, such as noise, causes Vo to increase. That will result in Vn increasing and now the voltage difference is less than 19.8 mV and so the output will drop. But what happens if it drops below 1.980 V? Then the voltage at Vn will drop, too, and the voltage difference will be greater than 19.8 mV and so the output voltage will go back up. This is the essence of negative feedback -- a departure of the output signal away from where it should be will result in the amplifier driving the output signal in the opposite direction of the departure.

Now consider that even poor opamps have open loop gains of 100,000 instead of just 100 and what you see is that the differential output voltage needed to saturate the output is on the order of 0.1 mV instead of 100 mV. Most modern run-of-the-mill opamps have gains in excess of one million, meaning that the differential voltage is going to be less than 10 microvolts in most circumstances (keep in mind we are ignoring other non-idealities such as input offset voltage). This is the basis for the assumption that a virtual short exists across the inputs of an opamp that is operating in its active region. In reality there must be some differential voltage in order to produce an output voltage (other than zero, of course), but that differential input voltage is so small that it is essentially zero.

Anhna
I see 2 different things in what you said. Please correct me if I'm wrong.

In #5 you said Vout cannot change Vin if there is no neg-fb. Ok I see that. Opamp + neg-fb.

In #6 you said Vin+ can only equal Vin- if there is neg-fb. But I thought "keeping V+ and V- equal was a feature of all opamps even without neg-fb?

Click to expand...

No, V+ and V- can only be the same with feedback. These are high resistance (high impendence) inputs, they are not outputs! How do you think the V- output ends up matching the V+ input voltage if it does not have access to the op amp output (changing) voltage in one of the standard amplifier configurations?

If I connect V- to ground and V+ to positive supply! how would you have thought they could be the same. Op amps are not magical. They do what you tell them to do with the circuity you add.

You can set up an op amp to allow it to stabilize itself (make V+ and V- become equal) with feedback, they don't do it without proper circuitry attached.

Four threads on op amp feedback - all active at the same time and you still don't have this basic concept down? You need to take some time to THINK and analyze what you've read and been told on these four threads - all you've done so far is memorize little snippets of information out of context. Maybe sit down, set up some test circuits, measure voltages and figure out how they work.

Also, a complete textbook on op amps would be good to carry you through little by little instead of expecting to learn everything from a single chapter of an online book.

Four threads on op amp feedback - all active at the same time and you still don't have this basic concept down? You need to take some time to THINK and analyze what you've read and been told on these four threads - all you've done so far is memorize little snippets of information out of context. Maybe sit down, set up some test circuits, measure voltages and figure out how they work.

Click to expand...

Hola Quique,

I agree, it is time to go that way; precisely what I did to start acquiring some familiarity with op amps.

No matter how much they are frowned upon, I keep one of my breadboards always handy to test basic (simple) things. Otherwise, solder on a small, neat perfboard.

To save you some unexpected/unexplained surprises, to carry out those tests, make sure you use a dual power supply. Later, do really learn how you polarize an opamp fed with a single supply.

Thanks to those of you responded with helpful explanations. I guess the others are geniuses who never got confused with anything in their lives.

For the record, I don't memorize things, the reason I'm going into so much detail is because I want to understand how they function. Otherwise I would have simply memorized the gain formulas for each and used those. I agree I should sit down and test to see what I get from that opamp.
I dont have access to an oscilloscope and I only have a TL071 that I got out of an old charge controller, so Ill have to sit down and test different input voltages and go step by step with my multimeter.

I see now where I was confused. The most helpful answer was dl324:

"You're confusing negative feedback with inverting operation."

And then anhnha nailed it with Jony's post. That was exactly the clarification I needed.

Yes, GopherT responded what many others have responded before. But Jony's post has something worth mentioning which is that it explains step by step how to get through the process and it specifically mentions how the voltage differential gets reduced. This line is vital:

You who have experience in this might take it for granted but its confusing to many others (otherwise opamps would be an easy topic and there wouldn't be so many questions about them posted everywhere). It might be a good idea to make that post a sticky for op amps topic.

Understanding how something works makes you tech-savvy, explaining how something works to others makes you a good teacher.

Understanding how something works makes you tech-savvy, explaining how something works to others makes you a good teacher.

Thanks again

Click to expand...

We all can and we all do teach quite well. However, when a "student" comes and says he wants to understand instead of memorize and does neither, then we get frustrated with the student.

Additionally, we are not just "tech savy" because we were born that way. We are tech savvy because we put time into our studies or time onto our own efforts. A student that just has the attitude of "learn me" is not going to learn, no matter how good his teachers are.

You are obviously free to keep asking questions here but you will advance much faster by reading a text book (hopefully one with a laboratory experiment section that you can use to prove each concept for yourself.

Thanks GopherT. Please don't assume Im not putting in the time or effort because I am. Im not an electronics engineer so I haven't put as much time into it as you obviously, I am a biochemical engineer. This is my hobby. I just set up my TL071 which is the only thing I have on hand. I connected both Vcc +/- to two 9V batteries to create a split supply from -9V to 9V. I set up a 10k pot coming from a 3.7V battery as Vin into the inverting input of the op amp. I gave it a negative feedback loop with two 1k Ohm resistors. But I have a question on the set up; should the GND from the 3.7V battery be connected to the GND of the 9V split supply?

Thanks GopherT. Please don't assume Im not putting in the time or effort because I am. Im not an electronics engineer so I haven't put as much time into it as you obviously, I am a biochemical engineer. This is my hobby. I just set up my TL071 which is the only thing I have on hand. I connected both Vcc +/- to two 9V batteries to create a split supply from -9V to 9V. I set up a 10k pot coming from a 3.7V battery as Vin into the inverting input of the op amp. I gave it a negative feedback loop with two 1k Ohm resistors. But I have a question on the set up; should the GND from the 3.7V battery be connected to the GND of the 9V split supply?

Im working on a picture of the setup...

Click to expand...

For a start, put (-) battery to ground between your two 9V.

Excellent start. By the way, I am not an EE, I have degrees in biology and chemistry. I am Not an engineer.

The explanation is that:
1. As Vin increases, it makes Vo greater.
2. As Vout gets greater (and since it is fed back to Vn (the inverting input), then a greater voltage is produced at this terminal.
3. As seen before, a greater voltage at Vn will tend to produce a smaller Vo. So this is why Vo goes down. Cause is greater Vn, effect is smaller Vo. This is just how the opamp works internally, which results in reducing Vo.

This is only for opamps configured with negative feedback? Iow, Vo will only go down when Vn is large, if there is negative feedback? Because I recall when starting opamps in the previous page that a large + voltage to Vn (inverting input) would generate a smaller Vo. So its not really a matter of the negative feedback, is it?

Click to expand...

Hi,

Vout will go down with Vn (the inverting input) because that input always causes Vout to go down when it goes higher in voltage. But the reason for negative feedback (rather than just applying a separate voltage) is to make the amplifier characteristics better.

One of the main advantages to negative feedback is accuracy of the output voltage with respect to the input voltage at (in this case) the non inverting input Vp. The negative feedback makes up for the deficiencies of the open loop gain which is not well specified and is not stable over temperature. The closed loop gain is highly predicable and stable, and that comes from negative feedback. The drawback is we loose some of the original gain in the process. Even so, it was probably one of the most important inventions in electronic history.

A better overall view of the op amp is that it is an "error" amplifier, and different circuit configurations have different advantages.